Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  CBSE XII  >>  Math  >>  Application of Integrals
0 votes

The area of the region bounded by the y-axis ,$y=\cos x$ and $y=\sin x,0\leq x\leq \large\frac{\pi}{2}$ is\begin{array}{1 1}(A)\;1\sqrt 2sq.units & (B)\;(\sqrt 2+1)sq.units\\(C)\;(\sqrt 2-1)sq.units & (D)\;(2\sqrt 2-1)sq.units \end{array}

Can you answer this question?

1 Answer

0 votes
  • The area enclosed by the curve $y=f(x)$,the $x$-axis and the ordinates $x=a$ and $x=b$ is given by $\int_a^b ydx.$
  • $\sin(A-B)=2\cos\large\frac{(A+B)}{2}$$\sin\large\frac{(A-B)}{2}$
  • Principle solution of $\sin\theta=n\pi\pm(-1)^n\alpha$
  • $\int \cos \theta=\sin\theta+c$
Step 1:
We have to find the area of the region bounded by the $y$-axis,$y=\cos x$ and $y=\sin x$.
Where $x$ ranges between $0\leq x\leq \large\frac{\pi}{2}$
First to find the area,let us find the point of intersection of $f(x)=\sin x$ and $g(x)=\cos x$
At the point of intersection $\sin x=\cos x$
$\Rightarrow \sin x=\sin(\large\frac{\pi}{2}$$-x$)=0
Now applying $\sin(A-B)=2\cos\large\frac{(A+B)}{2}$$\sin\large\frac{(A-B)}{2}$,we get
$\;\;=2\cos \big(\large\frac{x+\Large\frac{\pi}{2}-x}{2}\big)\normalsize\sin\big(\large\frac{x-(\Large\frac{\pi}{2}-x)}{2}\big)$$=0$
But $\cos\large\frac{\pi}{4}=\large\frac{1}{\sqrt 2}$
$\Rightarrow 2.\large\frac{1}{\sqrt 2}$$\sin(x-\large\frac{\pi}{4})$$=0$
$\sin(x-\large\frac{\pi}{4})$$=0\Rightarrow x-\large\frac{\pi}{4}$$=0$
Step 2:
Applying the principle solution we get,
$\Rightarrow x=\large\frac{\pi}{4},\frac{5\pi}{4},\frac{9\pi}{4}$........
But it is given $x$ values ranges between $0$ and $\large\frac{\pi}{2}$
Hence we take only $\large\frac{\pi}{4}$ and discard the 0 points.
Required area is the shaded portion shown in the fig.
The area of the shaded portion is given by $A=\int_0^{\Large\frac{\pi}{4}}(\sin x-\cos x)dx.$
Step 3:
On integrating we get,
$A=\begin{bmatrix}-\cos x-\sin x\end{bmatrix}_0^{\Large\frac{\pi}{4}}$
On applying the limits we get,
$A=[-(\cos\large\frac{\pi}{4}-$$\cos 0)-(\sin\large\frac{\pi}{4}$$-\sin 0)]$
But $\sin\large\frac{\pi}{4}=$$\cos\large\frac{\pi}{4}=\frac{1}{\sqrt 2},$$\cos 0=1$ and $\sin 0=0$
Substituting the values we get,
$\;\;\;=[-[\large\frac{1}{\sqrt 2}$$-1]-[\large\frac{1}{\sqrt 2}]]$
$\;\;\;=1-\large\frac{2}{\sqrt 2}=$$(1-\sqrt 2)$ or $\sqrt 2-1$
$A=(\sqrt 2-1)$sq.units.
Hence C is the correct option.
answered May 6, 2013 by sreemathi.v
edited May 6, 2013 by sreemathi.v

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App